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lisov135 [29]
3 years ago
12

You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

-turn to go home, pick it up and now after 20s you are driving south with a speed of 20 m/s. What is your displacement after 20s?
Physics
1 answer:
Readme [11.4K]3 years ago
8 0

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

d_1 = v*t_1

d_1 = 10*20 = 200 m

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

d_2 = v*t = 20*20 = 400 m

so the net displacement is given as

\vec d = \vec d_2 - \vec d_1

\vec d = 400 - 200 = 200 m

so it will be displaced by total displacement 200 m

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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
A block of mass
Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0
2 years ago
Coal burns in a furnace, producing light and heat. This reaction is
olasank [31]

The reaction is Exothermic

8 0
2 years ago
Read 2 more answers
A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus
Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone  = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0
3 years ago
.Need and actual answer not a guess
cricket20 [7]

Answer:

it's C (this is not a guess)

8 0
2 years ago
Read 2 more answers
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